Problem: Simplify and expand the following expression: $ \dfrac{1}{2k + 6}- \dfrac{5}{2k + 4}+ \dfrac{4k}{k^2 + 5k + 6} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{1}{2k + 6} = \dfrac{1}{2(k + 3)}$ We can factor a $2$ out of denominator in the second term: $ \dfrac{5}{2k + 4} = \dfrac{5}{2(k + 2)}$ We can factor the quadratic in the third term: $ \dfrac{4k}{k^2 + 5k + 6} = \dfrac{4k}{(k + 3)(k + 2)}$ Now we have: $ \dfrac{1}{2(k + 3)}- \dfrac{5}{2(k + 2)}+ \dfrac{4k}{(k + 3)(k + 2)} $ The least common multiple of the denominators is: $ 4(k + 3)(k + 2)$ In order to get the first term over $4(k + 3)(k + 2)$ , multiply by $\dfrac{2(k + 2)}{2(k + 2)}$ $ \dfrac{1}{2(k + 3)} \times \dfrac{2(k + 2)}{2(k + 2)} = \dfrac{2(k + 2)}{4(k + 3)(k + 2)} $ In order to get the second term over $4(k + 3)(k + 2)$ , multiply by $\dfrac{2(k + 3)}{2(k + 3)}$ $ \dfrac{5}{2(k + 2)} \times \dfrac{2(k + 3)}{2(k + 3)} = \dfrac{10(k + 3)}{4(k + 3)(k + 2)} $ In order to get the third term over $4(k + 3)(k + 2)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{4k}{(k + 3)(k + 2)} \times \dfrac{4}{4} = \dfrac{16k}{4(k + 3)(k + 2)} $ Now we have: $ \dfrac{2(k + 2)}{4(k + 3)(k + 2)} - \dfrac{10(k + 3)}{4(k + 3)(k + 2)} + \dfrac{16k}{4(k + 3)(k + 2)} $ $ = \dfrac{ 2(k + 2) - 10(k + 3) + 16k} {4(k + 3)(k + 2)} $ Expand: $ = \dfrac{2k + 4 - 10k - 30 + 16k}{4k^2 + 20k + 24} $ $ = \dfrac{8k - 26}{4k^2 + 20k + 24}$ Simplify: $ = \dfrac{4k - 13}{2k^2 + 10k + 12}$